Solve triangle, given angle A=97.77 deg, incircle radius r=3.58, area S=97.48

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kristjan

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given: angle A=97.77 degrees, radius of incircle=3.58 and area of triangle S=97.48. How can I find remaining angles and sides of triangle?
 
kristjan said:
given: angle A=97.77 degrees, radius of incircle=3.58 and area of triangle S=97.48. How can I find remaining angles and sides of triangle?
Click to expand...

The drawing is fairly simple, but in order to know what approach would be best for you, I'd want to know what you know.

I myself know, for example, that the area is the inradius times the semiperimeter (S = rs), so you could find the semiperimeter from the given data. There are also a couple different formulas for finding area given one angle and a couple sides; so you could find something like the product of the adjacent sides, and also (if you find the right formula) their sum. I found this approach by looking in a table of triangle formulas, because one of the formulas needed is one I didn't know.

If this is meant to be done using only facts you have learned in a course, perhaps you can list what you have been taught. If you are allowed to look up formulas, do so. Then let us know what ideas you have found. Keep in mind that our goal is to help you solve the problem, not just to do it for you, so we need to see your thinking.
 
I used S=rs, S=1/2*cbsinA and cosine theorem a2=b2+ c2 -2bc*cos(A).
Then maby solve system of equations bc=2S/sinA, a+b+c=S2r and a2=b2+c2-2bc*cos(A), but I´m sure there is some other easier way.
Yes drawing is fairly simple and I´m aware of properties of incenter but can´t conclude from the drawing anyting more.
 
kristjan said:
I used S=rs, S=1/2*cbsinA and cosine theorem a2=b2+ c2 -2bc*cos(A).
Then maby solve system of equations bc=2S/sinA, a+b+c=S2r and a2=b2+c2-2bc*cos(A), but I´m sure there is some other easier way.
Yes drawing is fairly simple and I´m aware of properties of incenter but can´t conclude from the drawing anyting more.
Click to expand...

I'll tell you that the third formula I used was on that relates the inradius to angle A without reference to the area. This is a natural thing to want, because you have S = rs (relating area and inradius) and S = bcsin(A)/2 (relating area and angle), so something relating inradius and angle seems useful. You can derive this formula for yourself by drawing that picture and considering either of the congruent right triangles at vertex A with opposite side r. Hint: the formula will involve angle A/2, specifically.

One you've done this, you can reduce things to two equations in b and c.

Your equation I marked in red is probably a typo.
 
Now I see congruent triangles and I write side a in terms of b and c. Solved it; thanks! :p
 
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